Centroid and second moment of area

By the end of this topic you will be able to

Find the centroid of a composite section by splitting it into simple rectangles.
Calculate the second moment of area about the centroidal axis using the parallel-axis theorem.
Explain why the centroidal axis is the neutral axis for bending.

Why section properties matter

How a beam, shaft or column responds to load depends not only on the material but on the shape of its cross-section. Two numbers do most of the work:

the centroid, the geometric centre of the area, which locates the neutral axis in bending; and
the second moment of area $I$ , which measures how far the area is spread from that axis. A bigger $I$ means a stiffer, stronger beam for the same material.

Centroid of a composite area

Split the shape into simple pieces (rectangles are enough for most sections), measure the centroid of each from a reference line, and take the area-weighted average:

\bar{y} = \frac{\sum A_i\, y_i}{\sum A_i}

Second moment of area

For a rectangle of width $b$ and depth $h$ , about its own horizontal centroidal axis:

I = \frac{bh^3}{12}

Note the cube: doubling the depth makes the section eight times as stiff.

To find $I$ for the whole section about its centroidal axis, shift each piece with the parallel-axis theorem:

I = \sum \left( I_i + A_i d_i^2 \right)

where $d_i$ is the distance from the centroid of piece $i$ to the centroid of the whole section.

Interactive modelPick a shape and change its dimensions. Watch the centroid and I update.

This interactive model needs JavaScript. The theory, worked example and problems on this page work without it.

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Worked example: a T-section

A T-section has a flange 120 mm wide and 20 mm thick on top of a web 20 mm wide and 100 mm tall. Find the centroid and $I$ about the horizontal centroidal axis.

Pieces (measured from the bottom of the web):

Part	$A$ (mm²)	$y_i$ (mm)	$A\,y_i$ (mm³)
Web 20 × 100	2000	50	100 000
Flange 120 × 20	2400	110	264 000
Sum	4400		364 000

\bar{y} = \frac{364\,000}{4400} = 82.7 \text{ mm}

Second moment. Web: $I_0 = 20\times100^3/12 = 1.667\times10^6$ , $d = 32.7$ mm, so $I = 1.667\times10^6 + 2000\times32.7^2 = 3.81\times10^6$ mm⁴. Flange: $I_0 = 120\times20^3/12 = 0.08\times10^6$ , $d = 27.3$ mm, so $I = 0.08\times10^6 + 2400\times27.3^2 = 1.87\times10^6$ mm⁴.

I = 3.81\times10^6 + 1.87\times10^6 = 5.67\times10^6 \text{ mm}^4

The model, set to this shape, shows the same values.

Practice problems

Find $I$ for a 100 mm wide, 200 mm deep rectangle about its horizontal centroidal axis.

Show answer

$I = 100\times200^3/12 = 66.7\times10^6$ mm⁴.
A rectangle is turned from 100 wide × 200 deep to 200 wide × 100 deep. By what factor does $I$ about the horizontal axis change?

Show answer

$I$ was $66.7\times10^6$ ; it becomes $200\times100^3/12 = 16.7\times10^6$ mm⁴, a quarter of the original. This is why joists are stood on edge.
For the T-section in the worked example, how far is the extreme bottom fibre from the centroid, and the extreme top fibre?

Show answer

Bottom: 82.7 mm. Top: $120 - 82.7 = 37.3$ mm. The section is not symmetric, so the stresses at top and bottom differ.
Use the model to compare an I-section (B = 120, t_f = 20, H_w = 100, t_w = 20) with the T-section. Which has the larger $I$ , and what is the area of each?

Show answer

The I-section: $I \approx 19.1\times10^6$ mm⁴, area 6800 mm². The T-section: $5.67\times10^6$ mm⁴, area 4400 mm². Adding the second flange raises $I$ by a factor of about 3.4 for 55% more area.

Key formulae

Centroid of a composite area: $\bar{y} = \frac{\sum A_i y_i}{\sum A_i}$
Second moment of a rectangle (about its own centroid): $I = \frac{bh^3}{12}$
Parallel-axis theorem: $I = \sum \left( I_i + A_i d_i^2 \right)$

Lecturer notes: suggested plan, activity and assessment

Suggested plan (45 min). 10 min: why shape matters (hold up a ruler flat then on edge); 10 min: centroid table method; 15 min: parallel-axis theorem and worked example; 10 min: model comparison of T and I.

Activity. Give each pair a different shape in the model and ask them to predict whether $\bar y$ will move up or down when the flange gets thicker.

Discussion. Why do steel beams have flanges?

Assessment idea. Ask for $\bar y$ and $I$ of an L-angle with given dimensions, with the model as a self-check.